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 September 13th, 2009, 05:54 PM #1 Newbie   Joined: Sep 2009 Posts: 1 Thanks: 0 how can these to be factor? 3x^2 - 7x + 3 and 2x^3 + 128 i have no idea how the first one can be factor... i dont think it can be factor... the second one, i went: 2[(x^3 + 64)] could this be answer? seems like i could go more steps September 14th, 2009, 06:02 AM #2 Senior Member   Joined: Mar 2009 Posts: 318 Thanks: 0 To learn how to factor quadratics (something they were supposed to have covered back in early algebra), try here. For the second one, you need now to apply the difference-of-cubes factoring formula. (Memorize this, along with the sum-of-cubes formula and the difference-of-squares formula. You will need all of these!) September 14th, 2009, 10:25 PM #3 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 We can possibly factor 3x� - 7x + 3 using irrational numbers. Let the given expression equal to 0, so 3x� - 7x + 3 = 0. Using quadratic formula for this equation, we arrive with roots of x = (7 � ?(13)) / 6. Since 3x� - 7x + 3 = 0 is equivalent to x� - (7/3)x + 1 = 0, it's clear to see that (x - (7 + ?(13)) / 6)(x - (7 - ?(13)) / 6) = 0, then it's obvious that 3(x - (7 + ?(13)) / 6)(x - (7 - ?(13)) / 6) = 3x� - 7x + 3 = 0. ? 3(x - (7 + ?(13)) / 6)(x - (7 - ?(13)) / 6). To factor the expression 2x� + 128 (or 2(x� + 4�)), use the suggested link given by stapel, we have 2(x + 4)(x� - 4x + 16). Tags factor Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Albert.Teng Algebra 7 June 16th, 2012 05:30 AM daigo Algebra 3 June 14th, 2012 06:38 PM HellBunny Algebra 3 February 18th, 2012 11:31 AM Eminem_Recovery Algebra 11 June 19th, 2011 09:50 PM profetas Algebra 3 August 25th, 2009 10:20 AM

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